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Have two LEDs back to back reversed and am driving from 120V 60Hz receptacle. My resistor is getting very hot to the touch. How do I cool it without making the LEDs dimmer?

LEDs are Radio Shack 5mm Round White LED (2-pack) part number 276-0017: intensity 7000mcd (typical), viewing angle 30 degress, FW current 25mA, FW supply 3.3V (typical), 3.6V (max) Thanks

Just load up your circuit in your question, go to Simulate, click Add Expression and type in "P(R1)" in the box. Run it. Yowza! 3 watts peak! That's a lotta heat.

Now, on the simulation output plot, right click on "P(R1)" in the legend, and click "Show average". Now you get the average power 1.540W. That's still a lot.

Since you asked you probably are just using a normal resistor, typically good for 0.125 or 0.250 W. So when you make it dissipate so much more it's going to get SCORCHING hot, probably even breaking down eventually. BAD.

So you can go find a $9.1k\Omega$ resistor rated for 1.5W or more of power dissipation. They do make power resistors, google it. They cost a bit more than a normal resistor but still under a buck.

Another approach would be to use multiple resistors so they each get a little bit of the heat. Here's your circuit but with four $2.2k\Omega$ resistors in series. That's $8.8k\Omega$, pretty close to your original value, but now each resistor could be a 0.5W resistor. Might be easier for you to find.

This one simulates and plots $P(R_1)$ which is the same as for any of the resistors. $P(R_1) = P(R_2) = P(R_3) = P(R_4)$

Thanks. I rebuilt this with 4 resistors 2.2k 1/2 watt and they're still warm but not crazy hot. Good enough!